Question

Factories $\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 1$​

Answer 1

Identities :-

• $\sf\purple{{(x+y)^2={\underline{\underline{\pmb{x^{2}+2xy+y^{2}}}}}}}$

• $\sf\purple{{(x+y)(x-y)={\underline{\underline{\pmb{x^{2}-y^{2}}}}}}}$

Solution :-

$\sf :\implies \red{\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 1}$

$\sf \: = \: \: \dfrac{ {x}^{8} + 1 + {x}^{4} }{ {x}^{4} }$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{8} + 2{x}^{4} + 1 - {x }^{4} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \green{ \sf \: [On \: adding \: and \: subtracting \: {x}^{4} }]$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {( {x}^{4} + 1) }^{2} - { ({x}^{2}) }^{2} \bigg)$

$\: \: \: \: \:$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {x}^{2} \bigg)$

$\: \: \: \: \: \: \: \: \:$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg( {x}^{4} + 1 + {2x}^{2} - {x}^{2} \bigg)$

$\: \: \: \: \: \: \: \: \: \: \: \green{ \sf \: [On \: adding \: and \: subtracting \: {x}^{2} ]}$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg)\bigg( {( {x}^{2} + 1) }^{2} - {(x)}^{2} \bigg)$

$\sf \: = \: \: \dfrac{1}{ {x}^{4} } \bigg( {x}^{4} + 1 - {x}^{2} \bigg) \bigg(( {x}^{2} + 1 + x)( {x}^{2} + 1 - x) \bigg)$

$\sf \: = \: \: \bigg( \dfrac{ {x}^{4} + 1 - {x}^{2} }{ {x}^{2} } \bigg) \bigg( \dfrac{ {x}^{2} + x + 1 }{x} \bigg) \bigg( \dfrac{ {x}^{2} - x + 1}{x} \bigg)$

$\sf\red{ \: = \:\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } - 1 \bigg) \bigg( x + \dfrac{1}{x} + 1 \bigg) \bigg(x + \dfrac{1}{x} - 1 \bigg)}$

• $\sf\red {Hence }$

$\sf :\implies \purple{\: {x}^{4} + \dfrac{1}{ {x}^{4} } + 1} = \sf \purple \:\purple {\bigg( {x}^{2} + \dfrac{1}{ {x}^{2} } - 1 \bigg) \bigg( x + \dfrac{1}{x} + 1 \bigg) \bigg(x + \dfrac{1}{x} - 1 \bigg)}$

⠀⠀⠀ ━━━━━━━━━━━━━━━━━━━⠀